property of ellipse

Question

In my book there is an statement for a ellipse.

Referring to the below ellipse:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

The statement is

The product of the length of the perpendicular segments from the focii on any tangent to the ellipse is $b^2$ and the feet of perpendicular lie on its auxiliary circle and the tangents at these feet to the auxiliary circle meet on the ordinate of P and that the locus of their point of intersection is similar to ellipse as that of the original one.

I could not understand the statement. Can anybody explain it to me with a DIAGRAM(FIGURE) in simple words.

That's still vague. Which parts of the statement are hard to understand? The auxiliary circle? The foot of a perpendicular? — J. M. ain't a mathematician, Dec 15 '16 at 11:23

Ng Chung Tak · Accepted Answer · 2017-01-08T13:57:56.553

4

Hints:

$\Delta ACD \sim \Delta BCE$

Let $(\cos \theta,\sin \theta)= \left( \dfrac{AD}{AC},\dfrac{CD}{AC} \right)= \left( \dfrac{BE}{BC},\dfrac{CE}{BC} \right)$ so that $AD\times BE=AC\times BC \cos^2 \theta$

$AC+BC=2a$

$AB=2\sqrt{a^2-b^2}$

$(DC+CE)^2+(AD-BE)^2=AB^2$

$F$ lies on the ellipse $\dfrac{x^2}{a^2}+\dfrac{b^2y^2}{a^4}=1$

Addendum:

For hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, there are similar properties except the locus gives the re-scaled conjugate hyperbola

$$\frac{x^2}{a^2}-\frac{b^2y^2}{a^4}=1$$

with their transverse axes in common instead of asymptotes.

edited Jan 08 '17 at 13:57

answered Dec 15 '16 at 23:58

Ng Chung Tak

19,693

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Red and blue ellipses are inversions with respect to green auxiliary circle.. Right? – Narasimham Dec 27 '16 at 19:54
@Narasimham Wow! Good observation that I missed it. – Ng Chung Tak Dec 27 '16 at 19:56
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The inversion is not radial but along $y; CF $ is parallel to $y-$ axis – Narasimham Dec 27 '16 at 20:01
Yes, what a wonderful property! – Ng Chung Tak Dec 27 '16 at 20:11
Could you please draw a similar diagram for hyperbola? – Mrigank Jan 08 '17 at 08:45
http://imgur.com/GXwuuUu properties are pretty similar – Mrigank Jan 08 '17 at 08:52
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@Mrigank Updated with the hyperbola case. – Ng Chung Tak Jan 08 '17 at 10:08
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Thanks, I tried too, has some more properties.. http://m.imgur.com/kGnrTgC – Mrigank Jan 08 '17 at 13:36
Yes, as pointed out from comment of @Narasimham – Ng Chung Tak Jan 08 '17 at 13:55

property of ellipse

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